Precalculus textbook problem (self-study): In the game of craps, there are two ways a player can win a pass line bet. The player wins immediately if two dice are rolled and their sum is 7 or 11 . If their sum is $4,5,6,8,9$, or 10 , the player can still win a pass line bet if this same number (called the point) is rolled again before a 7 is rolled. Find the probability that the player wins a pass line bet with a 4 on the first roll.
Textbook answer in back of book: 1/36
Question: How did they get this answer? Is my solution below right?
My solution: There are infinitely many ways to win after first rolling a sum of 4. The second roll could be a 4, for example. Or the second roll might not be 4 but it might be different than 7. And so on. So you could win on the second roll, the third roll, the fourth roll, and on and on.
The probability of rolling a sum of 4 is 3/36.
The probability of rolling a sum different than 4 and 7 is 27/36.
Winning on the second roll or on the third roll are independent events, and so on.
So the probability of rolling a sum of 4 first and then winning the pass line bet is
$$
\frac{3}{36}\cdot \sum_{i=0}^{\infty}\left[\left(\frac{27}{36}\right)^i\frac{3}{36}\right]= \frac{3}{36}\cdot\frac{3}{36}\cdot\frac{1}{1-\frac{27}{36}} = \frac{1}{36}
$$
Best Answer
Yes, your solution is correct. Here's another approach. Let $p$ be the probability that, after the first roll of $4$, the second $4$ appears before the first $7$. Then by conditioning on whether the next roll is $4$, $7$, or something else, we find that \begin{align} p &= \mathbb{P}[\text{$4$ before $7$}] \\ &= \mathbb{P}[4]\cdot \mathbb{P}[\text{$4$ before $7$}\mid4] + \mathbb{P}[7]\cdot \mathbb{P}[\text{$4$ before $7$}\mid7] + \mathbb{P}[\text{other}]\cdot \mathbb{P}[\text{$4$ before $7$}\mid \text{other}] \\ &= \frac{3}{36}\cdot 1 + \frac{6}{36}\cdot 0 + \frac{27}{36}\cdot p \end{align} yielding $p=\frac{3/36}{9/36}=\frac{1}{3}$. Hence the desired probability is $$\mathbb{P}[4]\cdot p=\frac{3}{36}\cdot\frac{1}{3}=\frac{1}{36}.$$